WAEC SSCE - Physics - 2010 (Essay)
Tambaya 1 Rahoto
(a) Write down the names of two particles used in explaining the wave nature of matter.
(b) State the wave characteristics which are exhibited by the particles named in (a) above.
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Bayanin Amsa
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Tambaya 2 Rahoto
An X-ray tube operates at a potential of 2500 V. If the power of the tube is 750 W, calculate the speed of the electron striking the target. [e = 1.6 x 10\(^{-19}\) C; mass of electron = 9.1 x 10\(^{-3}\) kg]
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Bayanin Amsa
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Tambaya 3 Rahoto
(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.
(b) Sketch a distance — time graph for a particle moving in a straight line with:
(i) uniform speed;
(ii) variable speed.
(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.
(ii) Sketch a velocity-time graph for the motion.
(a) Perfectly elastic and perfectly inelastic collisions
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both total linear momentum and total kinetic energy are conserved. | Total linear momentum is conserved, but kinetic energy decreases. |
| The objects separate after the collision. | The objects stick together after the collision and move with a common velocity. |
| No kinetic energy is converted into other forms such as heat, sound, or deformation. | This is the collision in which the maximum possible kinetic energy is converted into other forms. |
(b) Distance–time graphs
The gradient of a distance–time graph represents speed. Therefore, uniform speed gives a constant gradient, whereas variable speed gives a changing gradient.
(c)(i) Velocities at the ends of the 12 s intervals
For uniformly accelerated motion within each interval, the displacement is:
\[s=\frac{(u+v)}{2}t\]
Here, \(t=12\ \text{s}\). The final velocity of one interval is the initial velocity of the next interval.
-
First interval: \(u=0\), \(s=120\ \text{m}\).
\[120=\frac{(0+v_1)}{2}\times12\]
\[120=6v_1\]
\[v_1=20\ \text{m s}^{-1}\]
-
Second interval: \(u=20\ \text{m s}^{-1}\), \(s=300\ \text{m}\).
\[300=\frac{(20+v_2)}{2}\times12\]
\[300=6(20+v_2)\]
\[50=20+v_2\]
\[v_2=30\ \text{m s}^{-1}\]
-
Third interval: \(u=30\ \text{m s}^{-1}\), \(s=180\ \text{m}\).
\[180=\frac{(30+v_3)}{2}\times12\]
\[30=30+v_3\]
\[v_3=0\ \text{m s}^{-1}\]
| Time from start / s | Velocity / m s−1 |
|---|---|
| 0 | 0 |
| 12 | 20 |
| 24 | 30 |
| 36 | 0 |
(c)(ii) Velocity–time graph
Each 12 s interval has uniform acceleration, so each section of the velocity–time graph is a straight line. The area under each section equals the distance travelled in that interval.
Examination reminder: For successive intervals, do not restart from rest each time. Use the velocity at the end of one interval as the initial velocity for the next.
Bayanin Amsa
(a) Perfectly elastic and perfectly inelastic collisions
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both total linear momentum and total kinetic energy are conserved. | Total linear momentum is conserved, but kinetic energy decreases. |
| The objects separate after the collision. | The objects stick together after the collision and move with a common velocity. |
| No kinetic energy is converted into other forms such as heat, sound, or deformation. | This is the collision in which the maximum possible kinetic energy is converted into other forms. |
(b) Distance–time graphs
The gradient of a distance–time graph represents speed. Therefore, uniform speed gives a constant gradient, whereas variable speed gives a changing gradient.
(c)(i) Velocities at the ends of the 12 s intervals
For uniformly accelerated motion within each interval, the displacement is:
\[s=\frac{(u+v)}{2}t\]
Here, \(t=12\ \text{s}\). The final velocity of one interval is the initial velocity of the next interval.
-
First interval: \(u=0\), \(s=120\ \text{m}\).
\[120=\frac{(0+v_1)}{2}\times12\]
\[120=6v_1\]
\[v_1=20\ \text{m s}^{-1}\]
-
Second interval: \(u=20\ \text{m s}^{-1}\), \(s=300\ \text{m}\).
\[300=\frac{(20+v_2)}{2}\times12\]
\[300=6(20+v_2)\]
\[50=20+v_2\]
\[v_2=30\ \text{m s}^{-1}\]
-
Third interval: \(u=30\ \text{m s}^{-1}\), \(s=180\ \text{m}\).
\[180=\frac{(30+v_3)}{2}\times12\]
\[30=30+v_3\]
\[v_3=0\ \text{m s}^{-1}\]
| Time from start / s | Velocity / m s−1 |
|---|---|
| 0 | 0 |
| 12 | 20 |
| 24 | 30 |
| 36 | 0 |
(c)(ii) Velocity–time graph
Each 12 s interval has uniform acceleration, so each section of the velocity–time graph is a straight line. The area under each section equals the distance travelled in that interval.
Examination reminder: For successive intervals, do not restart from rest each time. Use the velocity at the end of one interval as the initial velocity for the next.
Tambaya 4 Rahoto
(a) Differentiate between plane polarization and interference as applied to waves.
(b) List two uses of polaroids.
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Bayanin Amsa
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Tambaya 5 Rahoto
State one reason each why cathode rays:
(a) are not electromagnetic waves;
(b) cast sharp shadows of objects in their path;
(c) can rotate a light paddle wheel inside a discharge tube.
(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
Bayanin Amsa
(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
Tambaya 6 Rahoto
(a) Explain the terms: (i) inertia; (ii) inertial mass.
(b) List three factors which affect the rate of evaporation of water in a pond.
(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.
(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\(^2\) kgm\(^{-3}\) and 1.30 x 10\(^3\) kgm\(^{-3}\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Bayanin Amsa
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Tambaya 7 Rahoto
A stone is projected vertically upward with a speed of 30ms\(^{1}\) from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms\(^-2\)]
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Bayanin Amsa
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Tambaya 8 Rahoto
(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.
(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.
(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.
(i) Determine the speed, v, of the waves.
Copy and complete the table given in your answer booklet.
(iii) Plot a graph with f on the vertical axis and \(\lambda ^{-1}\) on the horizontal axis.
(iv) What does the slope of the graph represent?
(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
Bayanin Amsa
(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
Tambaya 9 Rahoto
(a) List two examples each of substances with:
(i) low viscosity;
(ii) high viscosity.
(b) When is a liquid said to be viscostatic?
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Bayanin Amsa
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Tambaya 10 Rahoto
A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Bayanin Amsa
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Tambaya 11 Rahoto
When a lead-acid accumulator is fully charged, evolution of gases occurs at the electrodes. Name these gases and the respective electrodes at which thcy are given off.
When a lead-acid accumulator is fully charged it can accept no more charge, and the current then decomposes the water in the electrolyte, so gas bubbles are released:
- Hydrogen gas is given off at the cathode (the negative plate).
- Oxygen gas is given off at the anode (the positive plate).
This vigorous "gassing" is the sign that charging is complete.
Bayanin Amsa
When a lead-acid accumulator is fully charged it can accept no more charge, and the current then decomposes the water in the electrolyte, so gas bubbles are released:
- Hydrogen gas is given off at the cathode (the negative plate).
- Oxygen gas is given off at the anode (the positive plate).
This vigorous "gassing" is the sign that charging is complete.
Tambaya 12 Rahoto
(s) State the Heisenberg Uncertainty Principle.
(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.
(a) Heisenberg Uncertainty Principle
It is impossible to determine simultaneously and with perfect accuracy both the position and the momentum of a particle. The more precisely the position is known, the less precisely the momentum can be known, and vice versa.
(b) Mathematical expression
The product of the uncertainty in position \(\Delta x\) and the uncertainty in momentum \(\Delta p\) satisfies
\[ \Delta x \,\Delta p \;\geq\; \frac{h}{4\pi} \]so the constant is \(\dfrac{h}{4\pi}\), where \(h\) is Planck's constant. (Equivalently \(\dfrac{\hbar}{2}\).)
Bayanin Amsa
(a) Heisenberg Uncertainty Principle
It is impossible to determine simultaneously and with perfect accuracy both the position and the momentum of a particle. The more precisely the position is known, the less precisely the momentum can be known, and vice versa.
(b) Mathematical expression
The product of the uncertainty in position \(\Delta x\) and the uncertainty in momentum \(\Delta p\) satisfies
\[ \Delta x \,\Delta p \;\geq\; \frac{h}{4\pi} \]so the constant is \(\dfrac{h}{4\pi}\), where \(h\) is Planck's constant. (Equivalently \(\dfrac{\hbar}{2}\).)
Tambaya 13 Rahoto
(a) State two factors which affect the mass of elements deposited during electrolysis.
(b) List two non-electrolysis.
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Bayanin Amsa
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Tambaya 14 Rahoto
(a) State three conclusions that can be drawn from Rutherford's experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom
The diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from \(n_2\) to \(n_\theta\), calculate the:
(i) frequency;
(ii) wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\) Js; le V = .6 \(\times 10^{-19}\) J; C = 3.0 \(\times 10^8\) ms\(^{-1}\)]
(c) The following nuclear equations represent two types of radioactivity.
\({}^{226}_{88}R_a \to {}^{222}_{86}R_n + {}^a_2a\) (Equation A)
\({}^{14}_7N + {}^4_2a \to {}^{17}_8O + {}^1_1p\) (Equation B)
Identify each type and explain briefly the difference between them
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Bayanin Amsa
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Tambaya 15 Rahoto
(a)
When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.
(b) A proton moving with a speed of 5.0 x 10\(^{5}\) ms\(^{-1}\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\(^{-19}\) C]
(c)
The diagram above illustrates a 9.0 V battery of internal resistance 0.5 \(\Omega\) connected to two resistors of values 2.0 \(\Omega\) and R \(\Omega\). A\(_1\) A\(_2\) and A\(_3\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:
(i) equivalent resistance of the combined resistors 2.0 \(\Omega\) and R \(\Omega\);
(ii) currents through A\(_1\) and A\(_3\) ; (iii) value of R.
(a) Why the candle flame spreads out near the positive conductor
A candle flame is a region of hot, ionised gas: the burning gases contain positive ions, negative ions and free electrons. The pointed positively charged conductor sets up a strong electric field around itself, which exerts forces on these charges.
- Negative ions and electrons in the flame are attracted towards the positive conductor.
- Positive ions are repelled away from it.
The repelled positive ions collide with, and drag along, the surrounding neutral air molecules, producing a stream of moving air called an electric wind. This wind blows the flame gases outward, so the flame is pushed away and appears to spread out, exactly as shown in the diagram.
(b) Magnetic force on the proton
The magnetic force on a charge moving at an angle \(\theta\) to a field is
\[ F = qvB\sin\theta \]
With \(q = 1.6\times10^{-19}\,\text{C}\), \(v = 5.0\times10^{5}\,\text{ms}^{-1}\), \(B = 0.2\,\text{T}\) and \(\theta = 30^{\circ}\):
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(\sin 30^{\circ}) \]
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(0.5) = 8.0\times10^{-15}\,\text{N} \]
(c) The battery-and-resistors circuit
EMF \(E = 9.0\,\text{V}\), internal resistance \(r = 0.5\,\Omega\), and the main-line ammeter \(A_1\) reads the total current \(I = 4.0\,\text{A}\). The 2.0 \(\Omega\) and \(R\) resistors are joined in parallel.
(i) Equivalent resistance of the combined resistors
Total resistance of the circuit is
\[ R_{total} = \frac{E}{I} = \frac{9.0}{4.0} = 2.25\,\Omega \]
This total is the internal resistance in series with the parallel combination, so
\[ R_{eq} = R_{total} - r = 2.25 - 0.5 = 1.75\,\Omega \]
(ii) Currents through \(A_1\) and \(A_3\)
\(A_1\) is in the main line, so it reads the total current, \(I_1 = 4.0\,\text{A}\).
The p.d. across the parallel combination is
\[ V = I\,R_{eq} = 4.0 \times 1.75 = 7.0\,\text{V} \]
The current in the 2.0 \(\Omega\) branch (read by \(A_2\)) is
\[ I_2 = \frac{V}{2.0} = \frac{7.0}{2.0} = 3.5\,\text{A} \]
By Kirchhoff's current rule, the current through the \(R\) branch (ammeter \(A_3\)) is
\[ I_3 = I_1 - I_2 = 4.0 - 3.5 = 0.5\,\text{A} \]
(iii) Value of R
\[ R = \frac{V}{I_3} = \frac{7.0}{0.5} = 14\,\Omega \]
Check: \(\dfrac{1}{R_{eq}} = \dfrac{1}{2.0} + \dfrac{1}{14} = \dfrac{7+1}{14} = \dfrac{8}{14}\), giving \(R_{eq} = 1.75\,\Omega\), which agrees with part (i).
Bayanin Amsa
(a) Why the candle flame spreads out near the positive conductor
A candle flame is a region of hot, ionised gas: the burning gases contain positive ions, negative ions and free electrons. The pointed positively charged conductor sets up a strong electric field around itself, which exerts forces on these charges.
- Negative ions and electrons in the flame are attracted towards the positive conductor.
- Positive ions are repelled away from it.
The repelled positive ions collide with, and drag along, the surrounding neutral air molecules, producing a stream of moving air called an electric wind. This wind blows the flame gases outward, so the flame is pushed away and appears to spread out, exactly as shown in the diagram.
(b) Magnetic force on the proton
The magnetic force on a charge moving at an angle \(\theta\) to a field is
\[ F = qvB\sin\theta \]
With \(q = 1.6\times10^{-19}\,\text{C}\), \(v = 5.0\times10^{5}\,\text{ms}^{-1}\), \(B = 0.2\,\text{T}\) and \(\theta = 30^{\circ}\):
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(\sin 30^{\circ}) \]
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(0.5) = 8.0\times10^{-15}\,\text{N} \]
(c) The battery-and-resistors circuit
EMF \(E = 9.0\,\text{V}\), internal resistance \(r = 0.5\,\Omega\), and the main-line ammeter \(A_1\) reads the total current \(I = 4.0\,\text{A}\). The 2.0 \(\Omega\) and \(R\) resistors are joined in parallel.
(i) Equivalent resistance of the combined resistors
Total resistance of the circuit is
\[ R_{total} = \frac{E}{I} = \frac{9.0}{4.0} = 2.25\,\Omega \]
This total is the internal resistance in series with the parallel combination, so
\[ R_{eq} = R_{total} - r = 2.25 - 0.5 = 1.75\,\Omega \]
(ii) Currents through \(A_1\) and \(A_3\)
\(A_1\) is in the main line, so it reads the total current, \(I_1 = 4.0\,\text{A}\).
The p.d. across the parallel combination is
\[ V = I\,R_{eq} = 4.0 \times 1.75 = 7.0\,\text{V} \]
The current in the 2.0 \(\Omega\) branch (read by \(A_2\)) is
\[ I_2 = \frac{V}{2.0} = \frac{7.0}{2.0} = 3.5\,\text{A} \]
By Kirchhoff's current rule, the current through the \(R\) branch (ammeter \(A_3\)) is
\[ I_3 = I_1 - I_2 = 4.0 - 3.5 = 0.5\,\text{A} \]
(iii) Value of R
\[ R = \frac{V}{I_3} = \frac{7.0}{0.5} = 14\,\Omega \]
Check: \(\dfrac{1}{R_{eq}} = \dfrac{1}{2.0} + \dfrac{1}{14} = \dfrac{7+1}{14} = \dfrac{8}{14}\), giving \(R_{eq} = 1.75\,\Omega\), which agrees with part (i).
